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MECHANISM 


I N S R U C '1' ] O N I > A P E R 


PKEPARKI) BV 

Wai^tek H. James, S. B. 

f 

Instructor, Massachusetts Institute Technotogy 

AND 

IvAWRENCE S. Smith, S. B. 

f 

Instructor, Massachusetts Institute Technoi.ogy 


> > 

> . > 


AMERICAN SCHOOE OF C OR K F S P O N I) F N C IC 


CHICAGO 


U. S. A. 


ILLINOIS 













- ■ n r to ! ir w i n i**^ 

'jLliUKARY of OONBR£SS| 
j two Cfooies rtecMvoii | 

i OCTJZ. iWtt 


. I Slip 



Copyright 1908 by 

American School of Correspondence 


Entered at Stationers’ Hall, Condon 
All Rights Reserved 



0^-3bZ?p- 





MECHANISM 


The study of mechanism is the study of the laws that govern 
the motions and forces in machinery. Pure mechanism deals 
only with the amount and kind of motions, without regard to the 
forces transmitted, and in this instruction paper only this branch 
of the subject will be considered. That is, we shall-study how to 
proportion the parts to get the proper motions. Knowing this, the 
principles of machine design and strength of materials will teach 
us how large the parts must be to stand the stresses which come 
upon them. 

A Mechanism is a group of parts so shaped and arranged that 
a definite motion of one part will give other definite motions to 
the other parts. 

A riachine is a combination of mechanisms each of which may 
be doing a different kind of work, but the whole combination of 
mechanisms working together accomplishes some desired result. 
Take for example, a metal planer; the result which is to be ac¬ 
complished is the planing of a piece of metal, but in order to do 
this, several auxiliary results must be accomplished. The table or 
platen, to which the metal is fastened must be moved back and 
forth, the tool must be fed forward after each chip has been cut, 
and various other motions produced. There are certain pieces in 
the planer whose sole work is the moving of the platen, and all 
these pieces taken together form one mechanism; in like manner, 
there are certain parts whose sole work is feeding the tool forward 
each time, and all these parts taken together form another mechan¬ 
ism. All the mechanisms when brought together form the 
machine. 

MOTION. 

An object which is changing its position is said to be in 
motion and an object w^hich is not changing its position is at rest. 

Absolute and Relative notion. If a man is moving along in 
a sail boat wdthout changing his position in the boat, both man 
and boat have what is usually spoken of as absolute motion, that 



4 


MECHANISM. 


is, they are both changing their position with respect to the earth 
and .water around them. They both change their position in the 
same direction and at the same rate however, so that they are at 
rest with respect to each other. If now, the man slioiild start to 
walk forward in the boat, he would change his position with re¬ 
spect to surrounding objects faster than the boat and he would have 
motion relative to the boat. AYe thus have two kinds of motion, 
absolute and relative; absolute motion being the motion of a body 
with respect to some fixed object, and relative motion being the 
motion of one moving body with res])ect to another moving body. 




Since every object has more or less size and therefore can not 
be represented on paper as readily as can some particular point or 
line in the body, as far as possible in our study of the motions of 
bodies we shall consider points and lines, in place of the bodies 
themselves. ‘ 

Velocity is the rate of motion. If a bullet travels through the 
air at the rate of one hundred feet in a second, it has a velocity of 
one hundred feet per second. 

Direction of flotion. A point may have motion in a straight 
line, in a circle, or in any other curve. The tendency of any body 
which is in motion is to continue to move in a straight line unless 
caused to leave that path by some force applied to the body. 
The direction of the straight line in which a point is moving is 
called the direction of motion of the point. If an object is travel¬ 
ing in a circle, the direction of its motion at any given instant is 
the straight line tangent to the circle at the point where the object 









MECHANISM. 


5 


is at tlie given instant. For example, if the point A, Fig. 1, is 
moving aronnd the circle, the direction of its motion, when in the 
])Osition shown, is the line AB. In the same way in Fig. 2, if 
the point A is moving in the curved path, its direction when in the 
position indicated is the line AB. 


By choosing a convenient scale, as one inch equals one foot, 
or one inch equals ten feet, etc., depending upon how large the 
velocities are with which we have to deal, the line may he drawn 
of a length that will represent the velocity of the point A. Sup- 
])ose the circle. Fig. 1, to have a circumference of ten inches, and 
suppose the point A to be traveling around the circle at such a 
speed that it goes around onc*e in live seconds; then, since it travels 
ten inches in five seconds, it has a velocity of two inches per second. 
Accordingly,if we draw the line AB 
two inches long, tangent to the cir¬ 
cle at any given position of A, the 
line AB represents the direction and 
velocity of A at the instant under 
consideration. Of course, the direc¬ 
tion of motion of a point which is 
moving in any path other than a 
straight line is constantly changing, 
but the direction in which it is 



Fig. 3. 


moving at a given instant is the direction which it would take if 
tlie forces which constrain it to move in the curved’path were 
removed at that instant. 


Composite Motions. In Fig. 3, let A represent a body lying on 
a table; suppose AC and AD are strings making a right angle with 
each other. First, let the string AC be pulled with such force that 
it will cause A to move one inch in the direction AC, in one second; 
then A will be at B at the end of a second, the line AB repre- 
sejiting the direction and velocity of A when the string AC is 
j)ulled. Next, suppose the body to be back in its original position 
A, and* that the string A D is pulled with such force that it will 
cause A to move | of an inch in the direction AD in one second; 
then at the end of a second A will be at F, the line AE repre¬ 
senting the direction and velocity of the body when the string 
AD is pulled. 








6 


MECHANISM. 


Again, suppose botli strings to be pulled at tbe same time, 
with tbe same force as before. Then the pull on AC will still 
cause A to move toward C with a velocity AB, and the pull 
on AD will cause A to move toward I) with a velocity AE, 
so that at the end of a second the body will be at neither E nor B, 
but at some point II, whose position is at a distance BH from B, 
the line BII being ecpial and parallel to AE. The point II is also 
at a distance Eli from E, the line EH -being equal and parallel 
to AB. In other words, the path over which the body has 
moved is the line AH, which is the diagonal of a parallelogram 
whose sides are equal to AB and AE respectively. The lines 
AB and AE, which represent the velocities caused by the pulls 
on the respective strings, must be drawn to the same scale, that is, 
if the pull on AC causes a velocity of one inch per second and 



the pull on AD causes a velocity of inch per second, AE must 
be 4 as loim as AB.. 

4: O 

The velocities AB and AE are called eoinjjonent velocities^ 
and the velocity AH the resultani. 

Let us take another example. Suppose a ball is thrown 
toward the east with a velocity of ten feet per second, and the wind 

of live feet per 

second. Let us tind graphically how fast and in what direction the 
ball is actually moving. Using any convenient scale, draw the 
line AB, Eig. d, to represent ten feet, and the line AC to repre¬ 
sent five feet at the same scale, AC beincr at rio-ht arudes to AB 
since south is at right angles to east. Erom (J draw CD ])arallel 
to AB, and from B draw BD parallel to AC meeting CD at D; 
then AD, the diagonal of the rectangle thus formed, gives the 


carries the ball toward the south with a velocity 








MECHANISM. 


7 


direction and the velocity, at the same scale as before, of the actual 
motion of the ball. 

The preceding process is called composition of velocities; 
that is, if we know the velocity of a body in two directions at 
right angles to each other, we can find from these the real velocity 
and direction of the motion. Quite as frequently we have given 
the real direction and velocity and wish to find the velocity parallel 
and at right angles to some given line. Thus, suppose we know 
that a ball is moving at a speed of one hundred feet per minute 
toward the southeast and we wish to know how fast it is moving 
towards the east. Draw the line AD, Fig. 5, to represent one 
hundred feet, at a convenient scale; then draw the line AB, mak¬ 
ing with AD the same angle that a line running southeast makes 
with a line running east (that is, 45 degrees). From D draw DB 
perpendicular to AB, meeting A.B at B; the length of AB 
represents, at the same scale at which AD was drawn, the velocity 
of the ball in an easterly direction, and BD represents its velocity 
in a southern direction. This is called the resolution of veloci¬ 
ties, the velocity AD being resolved into two components, one in 
the direction AB, and one perpendicular to this direction. 

*BXAnPLES FOR PRACTICE. 

1. If a steamer travels 91,200 feet per hour, what is the 
velocity in feet per second? 

Ans. 25.83 feet. 

2. If the piston of an engine moves at the rate of 12.5 feet 
per second, what is the piston speed in feet per minute? 

Ans. 750 feet. 

3. A steamer is moving eastward at a speed of 800 feet per 
minute and the wind and tide carry it north at the rate of 100 feet 
per minute. Find gra])hically how fast and in what direction the 
steamer is actually moving. 

Ans. 810 feet per minute. 

4. A ball is thrown, with a velocity of 50 feet per second, 

across a stream 100 feet wide. The wind carries the.ball up stream 
10 feet per second. How far up stream from the point at which 
the ball was aimed will it strike ? Ans. 20 feet. 

*Note. Problems 3. 4, 5 and 6 are to be solved graphically. 




8 


MECHANISM. 


6. A platform is set up at an angle of 30 degrees witli the 
horizontal, and a weight is allowed to slide down the platform at a 
speed of ten feet per second. How long will it take the weight to 

get 10 feet below the point from which it started ? 

Ans. 2 seconds. 

6. The horizontal component of the velocity of a body is 20 
feet per minute. If the actual velocity is 52 feet per minute find 

graphically the vertical component. 

Ans. 45 feet per minute (about). 

7. A man is on a railway train which is moving at the rate 
of 25 miles per hour and walks toward the rear of the train at the 
rate of 88 feet per minute. How fast is the man actually moving ? 

Ans. 24 miles per hour. 

Temporary Center of Revolution. In Eig. 0, let All repre¬ 
sent a bar that forms some part of a mechanism, and suj)pose 

that for the moment the 
end B has a velocity in 

V 

the direction of and 
equal to BD, and that 
the end A has a veloc¬ 
ity in the direction of 
and equal to A(As the 
whole rod, for the instant, 
is moviiiP: as if it were 
revolvino;. about a cen- 
ter, the actual veloci¬ 
ties of the various points 
distances from the cen¬ 
ter. To find the center for the case shown, draw the lines 
AT and BT ])er])endicular to AC and BD, respectively, until they 
meet at T, which is the center about which the whole rod may be 
considered to be revolving. The velocities AC and BD cannot be 
chosen at random, but must be such that BE is equal to AE; 
fhat is, their components in the direction of AB must be equal. 

RFVOLVINQ BODIES. 

One of the most common motions which is found in machinery 
of all kinds is the motion of revolution, that is, the motion of a 


D 



in the rod are ])roportional to their 









MECHANISM. 


9 


body tuniing around a center or axis. The axis may pass through 
the body itself or it may lie entirely outside the body. Fig. 7 is 
an example of the first case, where the cylinder is revoking about 
the shaft A, which passes through the center of the cylinder. Fig. 
8 is an example of the second case where the block B is revolving 
about the shaft A. 

The speed of revolution of a body is generally described by 
pfivinor its number of revolutions per minute, which is abbreviated 
to B. P. M. 

We will use the following abbreviations throughout this in¬ 
struction paper: cZ^diameter; 'r=radius or distance of a point 
from the center about which it is revolving, and tt (pronounced 
Pi) = 3.U16. 

Let us consider a point C on the circumference of the cylin¬ 
der in Fig. 7. For every complete revolution of the cylinder the 




])oint C travels through the air a distance eipial to the circumfer¬ 
ence of the cylinder, or, since the circumference of a cylinder is 
3.141(1 times its diameter, the point 0 will travel 3.141() times 
the diameter of the cylinder. Therefore the actual velocity through 
the air or linear velocity as it is tenhed, of a point on the circum¬ 
ference of a revolving cylinder is ecj^nal to the diametei of cylindei 
times 3.141(1 times the number of revolutions in a unit of time. 

If we assume a foot as our unit of distance and a minute as our 
unit of time, the above principle may be stated as a formula: 
















10 


MECHANISM. 


Linear Velocity = tt d times R. P. M. (l) 

or “ =2 TT r times P. P. M. ( 2 ) 

This will give the linear velocity in feet per minute and to 
use this formula the diameter or radius must be expressed in feet 
or fractions of a foot. If d or r is in inches, the linear velocity will 
be in inches. If linear velocity per second is desired, the number 
of revolutions per second must be used instead of P. P. M. From 
this formula we see that the linear velocity varies directly as the 
diameter or radius, and also directly as the number of revolutions. 

The following examples will illustrate the use of the above 
formula: 

If a fly wheel is 10 feet in diameter and makes 75 revolu¬ 
tions per minute, how fast is a point on its circumference moving ? 
Psing formula (T), 

Linear Velocity = tt d X RdP. M. 

= 3.1416 X 10 X 75. 

= 2356.2 feet per minute. 

Therefore a point on the circumference of the fly wheel is 
moving at the rate of 2356.2 feet per minute. 

Suppose that the center of the block B, Fig. 8, is at a distance 
of 18 inches from the center of the shaft A and that the arm turns 
around the shaft 60 times per minute. To And the linear velocity 
of the center of block B, we must first reduce the 18 inches to feet. 
Then, 


Linear Velocity = 2 X 3.1416 X 1.5 X 60. 

= 565.49 feet per minute. 

The same formula may be used to find the diameter or radius 
when the linear velocity and revolutions are given, or to find the 
revolutions when the linear velocity and diameter or radius are 
given. To do this, substitute the known quantities in the equa¬ 
tion and solve by Algebra to find the unknown quantity. Let 
V Linear Velocity. 










MECHANISM. 


11 


Linear Velocity: 
Diameter: 

Revolutions: 


X ll. P. M. 



V 

77 X P. P. M. 


K. P. M. 


V 

77 d. 


EXAMPLES FOR PRACTICE. 


1. Find the linear velocity of a point on the rim of a fly 
wheel making 120 revolutions per minute. Assume fly wheel to 
be 7 feet in diameter. Ans. 2638.9 feet. 


2. How many turns will a 6-foot wheel make while going one 


mile ? 


Ans. 280. 


3. A locomotive is running at the rate of 40 miles per hour. 
How many revolutions are the 64-foot drivers making per minute ? 

Ans. 172. 


4. The drivers of a locomotive are 54 feet in diameter. The 
crank pin is twelve inches from the center of the wheel. Find 
the linear velocity in feet per second of the crank pin if the engine 
makes 25 miles per hour. Ans. 13.32. 


CYLINDERS AND CONES REVOLVING IN CONTACT. 

When two cylinders are arranged to revolve upon parallel axes 
which are at a distance apart just equal to the sum of the radii of the 
two cylinders, their circumferences will touch along one line, and 
if both cylinders revolve and we assume that there is no slipping 
of one surface on the other, the surface velocities of their circum¬ 
ferences must be equal, that is, the linear velocity of any point on 
the surface of one would be the same as the linear velocity of any 
point on the surface of the other. Then, if in Fig. 9, we let S 
equal the surface velocity of the cylinders, C the radius of cylin¬ 
der A, D the radius of P>, 41 the revolutions ])er minute of A, and 
N the revolutions j)er minute of P, we can obtain the following 
equations by substituting these values in formula (2):—- 


S = 2 TT 0 X M. 
S = 2 77 H X N. 










12 


MECHANISM. 


and dividing tlie first by the second we liave: 


S _^ 2 77 C X M CM 

^ ‘^2 ttD >rN irN 


therefore D N = 0 M or D X N = C X M. 

According to ratio and proportion in Algebra, 

M ___ D 
IT “C 



I 



Tills niay be stated as follows:— 

devolutions of Driver Hadiiis of Driven 
devolutions of Driven dadius of Driver • 

That is, the numbers of revolutions are to each other inversely as 
the radii, and therefore as the diameters. 

Two cylinders revolving in contact, without slipping, revolve 
in opposite directions; that is, if A revolves right-handed (like the 
hands of a clock), B will revolve left-handed. This is indicated by 
the arrows in Fig. 9. 

The following examples will illustrate the method )f calcula¬ 
tion for s])eeds of cylinders, according to the preceeding discus¬ 
sion. Suppose a wheel A, 2 feet in diameter, is revolving with its 
surface in contact with the surface of another wheel B. A makes 

































MEGHAN ISir 


13 


25 It. P. M. Assuming tliat no sli])|)ing occurs between the sur¬ 
faces of the two wheels, what must be the diameter of B in order 
that it shall make 75 II. P. ]\I. ‘i From the formula (^2) we have 

Diameter of B _ E. P. M. of 
Diameter of A E. P. M. of B 

01* Diametei of B_25 

2 feet 7 5 

Diameter ofBx75 = 2x25 


therefore Diameter of B = ^ feet = 8 inches. 

The same principles apply to cones which are revolving in 
contact. Let Fig. 10 represent the side view of two riglit cones 
which are in contact along the line EF, which is an element com¬ 
mon to both. Then, 


E. P. M. of A _ F G 
E. P. M. of B ~ FH 


(4 


That is, their revolutions 
are to each other inversely as 
the diameters of their bases. If 
the angles OEF and O'EF^ are 
known, instead of the diameters 
of the bases, the ratio of the 
number of revolutions may be 
found from the following^ trigonometric formula: 



E. P. M. of A Sine of angle OEF 

E. P. M. of B Sine of angle O'EF 


If the angle between the axes OE and O'E is given and it is 
desired to find the relative sizes of the two cones whose revolutions 
shall have a given ratio, the angles OEF and O’EF could be calcu¬ 
lated, but the calculation would be rather difficult and therefore a 
solution which is partly graphical is more convenient and usually 
sufficiently accurate. 

Let EO' and EO Fig. II, be the center lines of two shafts 
which lie in the same plane and make an angle ot 45 degrees 
with each other. The shafts are to be connected by two rolling 






















14 


MECHANISM. 


cones so that the shaft O'E shall-make 80 revolutions w'liile the 

shaft OE makes 40. The base of the lai*o;er cone is 12 indies, to 
» ^ ^ 
find the base of the smaller cone and the altitude of lioth. Since, 

from the preceding discussion, the sjieeds are inversely as the 

diameters of the cones, the larger cone must be on the shaft which 

is making the fewer number of revolutions, that is, O'E. 

Then from formula (4). 


Speed of O'E Diameter of base of cone on O E 
Speed of O E Diameter of base of cone on O'E 



Diameter of base of cone on OE 

12 


Solving the equation we get 

J)iameter of base of cone OE = 9 inches. 

Now draw the line PN parallel tc O'E and at a distance from 
O'E equal to the radius of the 
base of the cone O'E (in this case 
6 inches), at a convenient scale, 
and draw the line ML parallel to 
OE and at a distance equal 
to the radius of the base of the 
cone on OE (in this case 4^ 
inches). These lines intersect at 
the point C. A line from C to E 
is the element alonor which the 
cones touch each other. From C 
draw lines perpendicular to OE 
and O'E, meeting them at S and 
P respectively. Then SE and RE 
are the altitudes. To draw the cones, lay off ST equal to CS, and 
RV equal to OR. Join'E with T and Y and the cones are 
complete. 

CYLINDERS AND CONES CONNECTED BY BELTS. 

Frequently two shafts must be connected so that one may 
drive the other, and yet they are so far apart that cylinders cannot 



L P 

Fig. 11. 









MECHANISM. 


15 


be placed on tlieni wliicb. will touch each other. In this case it is 
customary to place on each shaft a cylinder, or pulley, and over 




these pulleys stretch a band, or belt made of some flexible mate¬ 
rial. The pulleys are fastened to the shafts so that the pulley and 
its shaft turn together, and the belt is stretched oxer the pulleys 
tic^ht enouo;h so that the friction between the belt and the surface 

o o 




linear velocity. A detailed study of the exact manner of finding 
the proper locations of the pulleys and drawing the pulleys and 
belts can be found in Mechanical Drawing Part Y, so that we will 










































































16 


MECHANISM. 


now study only the general ju’inciples involved, ])articularly with 
reference to making the necessary calculations. 

The ])rinciples which we learned above, concerning the rela¬ 
tive speeds of two cylinders in contact, a|)ply also to two ])ulleys 
connected by a belt. The rule governing the direction of rotation, 
however, is different. When the connection is made by means of 
a belt, the relation of the directions of rotation de])ends uj)on the 
way the belt is put around the pulleys. If the belt is ])ut on as 
shown in Fig. 12, known as open belt, the pulleys will turn in tho 





14. 



same direction. If the belt is as shown in Fitj. 13, known as a 
crossed belt, the ])ulleys will turn in o])posite directions. 

Crowning Pulleys. The outer surface of ])ulleys, instead of 
being made cylindrical is often made of larger diameter in the 
middle, as shown in Fig. II. A pulley having this increase of 
diameter in the middle is said to have a crowned face. The object 
of the crowning is to help the belt to remain on the pulley. A 
belt will always run to the j)art of the j)ulley which is largest in 
diameter; therefore, if the diameter increases towards the middle, 
the belt V7ill tend to run to the middle, and will be less likely to 
slip off. If the belt is to be at different places on the pulley at 













































MEC^HANISM 


17 


different times, as indicated by the full and dotted j)ositions in 
Fig. 15, the surface cannot he crowned, hut must he cylindrical. 

As problems similar to the following often arise, this one 
should he studied carefully until every step is understood. In 
Fig. 16, A is a shaft turning 60 It. P. M. in the direction indi¬ 
cated by the arrow. B and 0 are cylinders over which a strand of 
cloth is to pass as indicated, the cloth passing over the to]) of B 
and under the bottom of C. In order that J> and P may revolve 
in proper direction a crossed belt must he used between A and C, 



as shown. B is 24 inches in diameter, (■ is 18 inches in diameter, 
and the cloth is to travel at the rate of 120 feet per minute. E is 
a pulley 15 inches in diameter, which drives the ])ulley I), on the 
same shaft as B. F is a pulley 18 inches in diameter, driving 
juilley, G, on the same shaft as (2 The problem is to find the 
number of revolutions of B and (', and the diameters of I) and G. 

The circumference of B = 24 inches X tt — 2 feet X tt = 
6.2832 feet. Therefore, its surface velocity if it made one E. P. M. 
would he 6.2832 feet per minute. But since it is to have a surface 


































































18 


MECHANISM. 


velocity = 120 feet per minute (equal to that of the cloth) its 
speed must he 120 divided hy G.28d2 = 19.1 li. P. M. Eevolu- 
tions of C are found in the same way. 

18 inches = 1.5 feet. 1.5 feet X 3.1416=4.7124 feet = 
\3urface speed. 

120 divichid by 4.7124 = 25.4 E. P. M. 

Eow from formula (3): 

Eevs. of D _ Diameter of E 
Eevs. of E Diau^eter of D 

and since the revolutions of D = the revolutions of P 

19.1 15 

60 X 

19.1 a? = 15 X 60 
a?-=47.12+ 

therefore, pulley D is 47^ inches 
in diameter. 

Eevs. of G Diameter of F 
Eevs. of F Diameter of G 

25.4 _ m 

60 X 

25.4 = 18 X 60 

X = 42.5 (approx.) 

therefore, pulley G is 42.5 inches 
in diameter. 

Taper Cones. Suppose two equal frustums of cones are 
arranged on parallel shafts,' as shown in Fig. 17. Shaft A, turn^ 
ing at a constant speed, can be made to drive shaft B at different 
speeds by changing the location of the belt on the cones. In 
order to use cones with a straight taper, a crossed l)elt should be 
used. If an open belt is used, the cones should be shaped as 
shown in Fig. 18. It is necessary to give this curvature to the 
face in order to keep the same tension on the belt for different 
j)Ositions. With a crossed belt this condition is fulfilled by 
making the face a straight line. If the shafts are far apart, the 
straight cones are sometimes used for an open belt. 








































irECHANISM. 


19 


SiiK'o tlio (‘alciilations for finding tlio j)r()per shape of the 
cones for an oj)en l>elt are diliicnlt and are not often needed, we 
will give our attention only to the straight cones for a crossed 
belt. 

In Fig. 17 let shaft A be the driver, turnintr at a constant 
speed of 100 11. P. M. The small end of the cone on A is Iv 
inches in diameter. IVe wish to find 
the diameter of the laroe end of the 
cone on P, in order that the slowest 
sr>eed of B may be 75 P. P. M. Then 


if the cones are alike, to find the 


greatest speed of B: 

First 


Fi" 


18 . 


E. P. M. of A 
E. P. M. of B 

100 


KK 

ini 

EK 


Therefore If K 


1200 

i o 

Eow, if the cones are equal 

G L = E D 


75 EK 
10 inches. 


75 “ 12 
= 12 X 100 


and 

Then 


12 


F 0 = E K = 10. 
E. P. M. of A 
E. Jh M. of B 
100 ^ 12 
~ 10 


Ct L 
F"0 


X 

1000 


X 


Ir. 


= 1391 , 


§■ 


Therefore, the larger ends of the cones must be 10 inches in 
diameter and B will have a variation of S])eed from 75 E. P. to 
1881 E. P. M The length of -the cone does not affect the ex- 

o o 

treme speeds, but the longer the cones the more easily the inter¬ 
mediate s[)eeds may be obtained. 

Cones arranged as explained above are used to drive drying 
machines in bleacheries and similar ])laces, where it is desired to 
vary the sj)eed af the machine according to the weight of the cloth 
which is being dried. Heavy cloth reij^uires a longer time to dry 
than light cloth and consequently can not be passed through the 





















20 


MK('IIAKISi\r. 


iiiacbine as rapidly. Tapered cones are also found in various cot¬ 
ton machines. 

Stepped Cones, Stepped cones are often used in place of the 
tapered cones. These are really a series of pulleys side by side on 
a shaft, a similar series being placed on the other shaft with the 
large pulley on the second shaft in line with the small pulley on 
the first shaft. The arrangement is shown in big. lb. Tn practice 
the several ])ulleys which make U]) the step]^ed cones are cast to¬ 
gether, both cones u&ually being cast from one pattern, d he cal- 



Fig. 


19 . 


dilation for the diameters of the end steps are the same as for the 
end diameters for the cones (Fig. 17). Assuming that both cones 
are alike, the intermediate steps can be found as follows:— 

Decide how many intermediate, steps are desired, then draw 
the center line ST, and draw the lines AB and CD perpendicular 
to ST and at a distance apart equal to the width of the face of one 

step multiplied by the whole number of steps minus one. Make 

* 

CD equal in length to the diameter of the largest step and AB 
equal to the diameter of the smallest step. Draw AC and BD. 
Draw EF parallel to and at a distance from CD equal to the width 
of the face of one step. The distance between E and F, the ])oints 
where EP^ intersects AC and BD, will be the diameter of the sec- 



































































]V[EcirANis]\r 


‘’I 


ond step. Other ste])s may be found in the same manner. To use 
this metliod all ste])s must have tlie same width of face. It would 
not be accurate for an o|)en belt, althongh if the distance between 
the shafts are large comj)ared with the diameter of the cones, an 
open belt might not give trouble on cones designed in this way. 

Ste[)])ed cones are very commonly used for driving lathes and 
other machine tools. 

EXAMPLES FOR PRACTICE. 

1. A shaft making ,115 revolutions per minute has keyed 
to it a pulley 30 inches in diameter. If the ])ulley belted to the 
36-inch pulley is 26 inches in diameter, how many revolutions will 
it make per minute ? 

Ans. 160 revolutions, (nearly.) 

2. A pulley 12 inches in diameter makes 180 revolutions per 
minute. The other pulley connected l)y belt to the 12-inch pulley 
should make 75 revolutions. How large must it be ? 

Ans. 28.8 inches in diameter. 

3. A motor shaft having a pulley 6 inches in diamete^’ makes 
840 revolutions. If this speed is to be reduced to 320 revolutions, 
what size pulley should be used ? 

Ans. 15| inches in diameter. 

Disk and Roller. Fig. 20 show^s a device by means of which 
one shaft turning always in the same direction with a' constant speed 
may drive another shaft, which is at right angles to it, in either 
direction and at different speeds. A is a disk on the end of the 
overhanging shaft S. On the shaft T is placed a roller 13 free to 
be moved up or down on T, but oii a key so that it must turn with 
the shaft. Some sort of a shipper must be ])rovided to hold 13 in 
the desired j)osition on T. AVdien one shaft turns it drives the 
other by means of the friction between A and B. If shaft S is the 
driver, turning in the direction indicated by the arrow, it will 
cause T to turn in the direction indicated by the lower arrow. If 
it is desired to have T turn in the opposite direction, keeping the 
direction of S the same, 13 must be moved up above the center line 
^)f S as shown dotted. If OH is the distance from the center of 
disk A to the point where the center line of the face of 13 strikes 







MEClTAISriSM. 


90 


A, and LR is the radius of B, we have the following equation,— 


ih r. ]\r. of B _ o B 

Jhl\]\I.of A “ L K 



from which, knowing the speed of one, the speed of the other may 
be calculated. It will be seen from this equation that if OB is 
decreased, that is, if B is moved nearer the center line of S, the speed 
of B will decrease if A is the driver, or the speed of A will increase 
if B is the driver. 




This device is used for driving drying machines where the 
cloth goes from another machine direct to the dj’ying machine, and it 
is essential that the speed of the two should be adjusted to a nicety, 
in order that excessive strain may not come upon the cloth. It 
is also used for driving the feed mechanism on machine tools, par¬ 
ticularly drills. 

Example: Suppose the disk A makes 70 revolutions per 
minute and disk B 124 revolutions. If OB is 18 inches what is 
the diameter of B ? ' 

B. P. M. of B O B 124 18 

li. r. M. of A ~ L li ~LTl 

124 X L r = 70 X 18 
,_70xl8 

= 10.10 inches (about). 





















































MECHANISM. 



Diameter would be 2 X 10.10 = 20.32 indies or 203 ''’^ incbes. 

EXAMPLE FOR PRACTICE. 

1 . 1 ) isc B, Fig. 20 , is 12 indies in diameter and makes 280 

revolutions per minute. If shaft S is to make 48 revolutions per 
minute how far from the center must the circumference of B be 
placed ? Ans. 35 inches. 

DISCONNECTING AND REVERSING HECHANISMS. 

Tight and Loose Pulleys. Fig. 21 shows a very common 
arrangement of pulleys, whereby the shaft A maybe stopped while 
the driving shaft B continues to run. Pulley C is keyed to 



Fig. 21. 

shaft A, while pulley D is free to run on A. Pulley E has its face 
e(]ual in width to the sum of the other two and is keyed to B. 
AVhen the belt is in the position shown in full lines, D turns with 
E, while shaft A remains at rest. The belt is guided by a shipper 
and when it is desired to start A, the belt is moved into the posi¬ 
tion shown by the dotted lines. 

Jaw Clutch. Fig. 22 illustrates an arrangement by which 
the gear, or pulley, A may be attached to turn with the shaft, or 
disconnected so as to turn independently of the shaft. The hub 
of A has jaws or teeth on its end, and the sliding piece B has simi¬ 
lar teeth, the projections on B htting into the indentations of A. 
The shaft may turn freely in A while B slides on a key so that it 
must turn with the shaft. MTien B is in the position shown, the 
shaft and A may turn independently of each other, but when B is 

































24 


MECHANISM. 


moved to the left so that its teeth engage witli the teeth on the 
hub of A, it serves to elani[) A to the shaft. The principle here 
involved is used in many ditfei’ent forms of clutches. 

Friction Clutches. A jaw clutch such as illustrated in big* 
22, if thrown in while the shaft or gear is turning brings a sudden 
shock on the parts, which is likely to cause damage, especially if 
the speed is high and much power is transmitted. (jonse(piently, 
a clutch known as a friction clutch is often used, which, as its 
action depends n])on the friction between two parts, will impart 
motion more gradually thereby avoiding the shock. Eig* 23 illus¬ 



trates a simple clutch of this kind. E is a ])iece free to slide on a 
key, and held in position by a shipper. A is a pulley or gear loose 
on the shaft. The outer surface of B and the inner surface of A 
are tapered to tit each other. AVhen B is thrown to the left it 
clamps to the shaft in the same manner as the jaw clutch. 

Two Clutches for Reversing. Fig. 24 shows how two clutches 
can be arranged so that either one may he thrown in, thus giving 
motion to the shaft in either direction. The clutches are similar 
in principle to the one shown in Fig. 23. The left-hand clutch is 
shown partly cross-sectioned. The ])iece A is attached to a shipper 
which when moved to the right throws in the right-hand clutch; 
when moved to the left throws in the left-hand clutch^ H the puF 





































MECHANISM. 


25 


ley of the ri^lit-hand clutch is driven by an open belt and the pul¬ 
ley of the left-hand clutch by a crossed belt, the shaft will be driven 



Fig. 23. 


in either direction according as one or the other clutch is thrown 
in. AV hen A is in the position shown neither clutch is in and the 
shaft is at rest. 

SCREWS. 


The Hel ix. Since all screws depend njion a curve known as 



Fig. 24. 



the helix, it will be necessary to understand soinetliing about this 
curve before attempting the study of the screw. 

Take a cylindrical piece of wood, such as is shown in Fig. 2b, 
and a rectangular piece of paper, ABP^E, with the side AB equal. 



























































































26 


MECHANISM 


to the circumference of the cylinder, and the side AE equal to the 
length of the cylinder. Now, if we lay off along AE a distance 




B 


C 

J 

F 


AD, and draw the line DC parallel to AB, we have the rectangle, 
ABCD. Now draw the diagonal, AC, of the rectangle and wrap 
the paper around the cylinder, keeping the side AE on an element 
of the cylinder; the paper will just cover the cylinder, the edge 


h-PITCH- - —>1 

I A|23456789I0IM2I B 



BE meeting the edge AE. The point C will coincide with D and 
lie on the same element of the cylinder as A and at a distance from 
A equal to AD. The shape which the line AC now takes is called 
a helix, and the distance AD is called the pitch of the helix. If 



























MECHANISM 



we choose another point, H, half-way between A and D, and draw 
a line ILJ parallel to AC, this line when wrapped around the cyl¬ 
inder will form another helix of the same pitch as the first one. 
The two together form what may be called a double helix. Tf the 
line AD is divided into three equal parts instead of two, and lines 


A R B 



parallel to AC be drawn from each of the points of division, we 
would have on the cylinder three parallel and equidistant helices 
of the same pitch, or a triple helix. Fig. 26 shows a single right- 
hand helix. Fig. 27 a single left-hand helix. Fig. 28 a double right- 
hand helix, and Fig. 29 a triple right-hand helix. 

If a helical groove is cut on the surface of a cylinder, a screw 


-PITCH- 



thread is formed, the thread being the material which is left be- 
tween the successive turns of the helical groove. A cylinder 
having a thread on it is called a screw, and a ])iece having a cyl¬ 
indrical hole in it, with a thread around the hole, is called a nut. 
The groove which is cut to form the thread may be in the form of 
a single, double or triple helix, and either right-hand or left-hand; 
the thread would be described in the same terms. 





















MECHANISM 




Eig. 80 sllo^ys a right-hand and a left-hand single A" thread 
and a right-hand and a left-hand single square thread. 


i<-PITCH->. 



One complete turn of a screw causes it to travel through its 
nut a distance equal to its pitch. 



KlOHT-HAI^D THREAD, • KIGHT-HAND THREAD. 

Vi^. 30. 


There are certain principles which apply and certain problems 
which arise, in connection with all forms of screws, and we shall 
now consider some of the most impoidant of these. 

Let Fig. 31 represent a jack screw. The jiitcli of the screw 
is V inches; the length of the bar from the axis of the screw to 







































MECHANISM 


29 


the point JR, where the force is applied, is A inches. A force of 
F pounds is exerted at It, raising the weight W. 

The following equation holds true, 

F Distance W moves 




Distance I’ moves 


7 ) 


Suppose the screw to ])e turned once; then F will move 
through a distance approxiinately etpial to the circumference of a 
circle whose radius is A; that is, F moves a distance 2 tt A. In 
turnino; the screw once, is raised a distance P. Then, suhsti- 
tilting, in equation (7), these quantities for the distances which I 
and AT respectively move, we get 


F 


P 


'I- 


A 


-.L__ 


133- 


bar 


R 


P= PITCH 




I I >1 
ii 


2 'H- A F = AT X p 

ATxP = 27rAxF 

From this equation, if any 
three of the quantities F, AT, 

P or A are known, the fourth 
may be found. P and A must 
both be expressed in feet or 
both in inches. 

For example, let P = ^ inch, A = 5 feet, F = 100 pounds. 
Substituting in the formule we get 


Fiv. 31. 


100 




AT “ 2 X 3.1416 X 60 

.25 AT -= 37699.2 

AT = 150,797 pounds (nearly). 


EXAMPLES FOR PRACTICE. 

1. Suppose a weight of 200,000 pounds is to be raised by a 
jack screw. The jiitch is .20 inch, and the force applied is 12o 

pounds. AThat must be the length of the bar? 

Ans. 51 inches (nearly). 

2. AATiat should be the pitch of the screw of example 1 if 

the bar is made (> feet long? Ans. .28-]- inches. 





































80 


MECHANISM. 


8. A jack screw lias tlie following diiriensions: Tx^iigth of 
arm, 51 feet; pitch, .125 inch. If 20(),00() pounds are to be raised 
what force must he applied at the end of the bar? 

Ans. 00 pounds. 

Combination of Screws. Fig. 82 illustrates a method by 
which the space moved through by the nut C may be made very 
small in comparison with the space moved through by a point on 
the circumference of the hand wheel, without using a very fine 
pitch on the screw. Great pressure can thus be produced at C by 
a small force on the wheel, wdthout danger of stripping the thread, 
which might occur if the screw had a fine pitch. The upper part 
of the screw E has a coarse thread which acts in the nut B, this 
nut being rigid with the frame I). The lower part of the screw 
has a pitch different from the upper part and acts in the nut C. 
C is free to slide up or down in the frame, but is prevented from 
turning by the lips L. 

Let P — the pitch of the upper part of the screw and p = the 
pitch of the lower part, both threads being right-hand, and P being 
greater than If the screw were attached to C by a collar in¬ 
stead of heincr threaded into it, one turn of the screw rierht-handed 

O' O 

in nut B would cause the screw to advance a distance P, carrying 
C forward the same distance. But since by the same turn the 
screw has drawn nut C on to itself a distance the actual distance, 
K, which C has advanced is 

K = p (9) 

If p were greater than P, K would have a minus value; that 
is, C would actually move u]) for a right-lianded turn of the screw. 
If one of the screws is left-handed and the other ri^ht-handed, the 
equation would become 

Iv = P+yi. (^lO) 

Let P = inch and j) — inch, and let both screws be right- 
handed. If the radius A of the hand wheel is 12 inches, how 
much ])ressure can be exerted at 0 by a force of 50 pounds, ap¬ 
plied to the circumference of the wheel? 

The same principle holds here as in the case of a simple 
screw; that is, the forces exerted are inversely as the distances 
moved through, or. 




MECirANISM. 


81 


Force at wheel Travel of () 

Force at C Travel of point on circuin. of wheel 



so that we must first find the distance C moves for one turn of the 
wheel. Substituting the values of P and ]) in formula (9), we get 

K = ^ = distance C moves for one turn of the 

screw. 


The distance which a point on circumference of wheel moveE' 



Fig. 32. 


— 2 77 A = 2 X 3.1416 X 12 = 75.4 inches, 
values in formula (11) we get 

50 


1 (5 


Force at O 75.4 


Sul)stitutino; these 


Solving, we get, force at C = 60,320 pounds. 












































32 


MECHAKISM. 


EXAMPLE FOR PRACTICE. 

1. Suppose the press shown in Fig. 32 has the following 
dimensions: P = J inch,= J inch, and the force at C is to be 
78,000 pounds. What force must be applied to the circumference 

of the wheel if it is 14 inches in diameter? 

A ns. 00 pounds ( nearly). 

Worm and Wheel. The mechanism known as worm and 
wheel consists of a wheel, as A, Fi(^. 33, havino; teeth on its cir- 
cumference, and a screw or worm W, meshing with the teeth on 
the wheel. The axis on which the wheel is fixed is usually at 
rio'ht ano-les to the axis of the worm. If the worm is turned, tlij 
action between it and the teeth on the wheel is similar to the 
action between a screw and the thread in a nut, so that the wheel 
will turn also. If the worm is single-threaded, one turn of the 
worm will cause the wheel to advance one tooth, so that in order 
for the wheel to make one com|)lete revolution, as many turns of 
the worm will be re(|uired as there are teeth in the wheel. If the 
worm is double-threaded, half as many turns will be required as 
there are teeth in the wheel; if the worm is triple-threaded, one- 
third as many turns will be required as there are teeth in the 
wheel. 

Suppose the worm be single-threaded and turned by a crank 
whose length is P; let the worm wheel have n teeth and have on 
its shaft a cylinder of radius K, with a cord wound around it sup¬ 
porting the weight L. If a force of F pounds is applied to the 
crank, the same principle holds true as in the other mechanisms 
which we have studied, namely, 

F Distance L moves , 

( 13 ) 


L 


Distance F moves 


If the worm makes one revolution, F will move throimh a 

I 

distance 2 tt K. The wheel makes — revolutions, and conse- 

II 

quently the cylinder to which L is attached will make - rev¬ 
olutions. If the cylinder made a whole revolution it would 
raise L a distance ecpial to its circumference = 2 K, therefore, 

when it makes one revolution it wdll raise L a distance ^ ^ ? Sub- 

n 

stituting this value for the distance L moves, in formula (12), and 
substituting for distance F moves, the quantity 2 tt we (ret 









:^[EC1TAKJS1L 


33 


2 TT IC 

Y _ __ 2 TT K 

\j 2 77 K 2 77 //K 


Tlierefore, 


Y K 

L“ y/U 



H the worm is douhle-tlireaded, L will l)e raised twice as far 
for one turn of the worm, so tliat the above formula would become 
for a double threaded worm and wheel, 

F _ 2 K 

L n Iv 

and for a tri])le-threaded worm and wheel, 

F __ 3 K 
L “ yy K 


Zi — 


Fig. 33. 




For example, a siugle-threaded worm makes 180 revolutions 
per minute and the worm wheel has 30 teeth, how many feet ])er 
minute will a weight be raised if the cylinder on the shaft of the 
worm wheel is 12 inches in diameter? 

For each revolution of the worm wheel the worm must make 
30 revolutions, hence the worm wheel makes 6 revolutions per 
minute. The circumference of the cylinder is 2 tt R or 2 X 
3.1416 X 6 = 37.699 inches and 6 X 37.699 ^ 226.197 inches 
= 18.85 feet. 











































34 


MECIlA^sISM 


EXAMPLES FOR PRACTICE. 

1. A crank of 18 inches radius is attacluMl to a worm of a 
single thread. The worm wlieel lias 48 teeth and is attached to a 
drum 14 inches in diameter. What weight can he raised if 80 
pounds is applied to the crank ? 

Ans. 9874 pounds. 

2. If the crank in above device is given 216 revolutions per 
minute how fast will the weight rise ? 

Ans. 16J feet (nearly). 

3. llow fast would it rise if the worm has a double thread ? 

Answer 33 feet. 

CAMS. 

A cam is a plate or cylinder having a curved outline, or a 
curved groove in it, which, by its rotation about a fixed axis, im¬ 
parts a backward and forward motion to a piece in contact with it. 
The motion imparted to the follower may be at right angles to or 
parallel with the axis of the cam, or at some other angle. 

The motions which cams are designed to give to their followers 
commonly belonged to one of the following kinds: uniform motion, 
harmonic motion, or “gravity” motion. The cam may give one 
kind of motion all the time or it may give different kinds of mo¬ 
tion at different stages in its revolution. 

o 

Before studying the cams themselves, we will consider the 
three kinds of motions mentioned. 

Uniform Motion. If a body moves through equal spaces in 
equal intervals of time, it is said to have uniform motion, that is, 
its velocity is constant. 

Harmonic Motion. If a point as C, Fig. 34, travels around 
the circumference of a circle with uniform velocity, and another 
point, as E, travels across the diameter at the same time at such a 
velocity that it is always at the point where a ])erpendicular let fall 
from C would meet the diameter, the point E would be said to 
have harmonic motion. Its velocity will increase from the start¬ 
ing point A until it reaches the center O, and from O to D its 
velocity gradually decereases to zero at D. 

Gravity Motion, or uniformly accelerated and retarded mo¬ 
tion, is also a motion where the velocity gradually increases until 
it reaches a maximum at the middle of the path and from there 





MECHANISM. 


35 


gradually decreases to the end. The rate of increase and decrease, 
however, is different from tliat in harmonic motion, the velocity 
being increased in gravity motion, by e(jnal amounts in e<|ual in¬ 
tervals of time, the spaces traveled over in successive intervals of 
time being in the ratio of 1, 3, 5, 7, etc., to the middle of the path, 
and decreasiiiDr in the same ratio to the end. 

The kind of motion which the follower is desired to have must 
he taken into consideration when designino- a cam. 

Plate Cams. Fig. 35 shows a plate cam designed to raise 
and lower the follower P with uniform motion. The point of 
is raised from A to e while the cam turns right-handed through 
two-thirds of a revolution, and is lowered again to its original posi¬ 
tion durinof the remainino- one-third of a revolution of the cam. In 
this cam the line of motion to the point of P passes through the 
point O, which is the center of the shaft on which the cam is fixed. 


B 



Fig. 34. 

To design such a cam, draw a circle with O as a center and 
passing through A. Since the raising of P is to take place during 
two-thirds of a revolution, right-handed, find a point E on the cir¬ 
cumference of the circle, such that the arc AIIE shall be ^ of the 
whole circumference, thus leaving arc ACE ecpial to g of the cir¬ 
cumference. Divide the arc AIIE into any number of equal parts 
and draw the radial lines DI, OCt, etc. Divide A^? into the same 
number of equal parts and from the points of division 1, 2, 3, etc., 
thus found, swing arcs around the center O, cutting 01, OG, etc., 
at pointsy, etc., respectively. A smooth curve dia\\n thiough 
A, /' y, etc., to m, will give the outline of that part of the cam 
which raises P. The outline of the rest of the cam is found in a 
similar manner, by dividing arc ACE into any number of equal 















30 


MECIIAXISM. 


parts and making a new division of A.e into tlie same number of 
equal parts, (points e, h.) 

Fig. 36 shows a cam which gives the same motion to the fol¬ 
lower as was given by the cam shown in Fig. 35, but in this case 
the line of motion of P, instead of passing through O, passes a dis¬ 
tance Oa to one side of O. The only difference in the method of 
desio-nino; this cam from that previously described is that instead 

o Jo 

of drawing radial lines through ])oints A, F, G, etc., we draw lines 


IX 




through these points tangent to a circle drawn with center O and 
radius Oa.' 

If in Figs. 35 and 30 it had been re(piired to raise P with 
harmonic motion instead of uniform motion, the only difference in 
construction would have been in makino^ the divisions of Ke. 
These divisions, instead of being equal, would be found as shown 
on a larger scale in Fig. 3T. On the line Ae as a diameter, draw 
a semicircle and divide this semicircle into as many equal parts as 
we divided the arc AIIE in Figs. 35 and 36. From the points 








































MECHANISM. 


37 


BCD, etc. (Eig. 37), draw perpendiculars to line A<^, cutting it 
at points 1, 2, 3, etc. The division A-1, 1-2, etc., thus found, 
are the divisions of A.e which would be used in tinding the cam 
outline, in place of the equal divisions which were used for uniform 
motion. 

Eig. 38 shows how A^ would be divided if the motion were 



to be ‘‘ Gravity.” Since A^^ is to be divided into as many equal 
parts as arc AHE, in this case it will be divided into eight parts. 
Now if the motion which the cam is to give to piece F is to be 
uniformly accelerated and retarded (that is, gravity motion), Ke 
must be divided in such a way that the distance from 1 to 2 is 
three times the distance from A'to 1; the distance from 2 to 3 is 
live times the distance from A to 1; 3 to 4 is seven times A to 
































38 


MECHANISM. 


1; 4 to 5 is seven times A to 1; 5 to 0 is five times A to 1; 6 to 7 
is three times A to 1; and 7 to ^ is equal to A to 1. Therefore 
the whole line Ae is l-l-3d-5-]-7 + 7-l-5-|-3 + l or 32 times the 
distance from A to 1, or in other words, A-1 is of the whole 
line A^, and 1 — 2 is of Ae, and so on. To.divide Ae so that 
the divisions may hear this relation to each other, draw the line 
AH at any convenient angle, and choosing any convenient distance 
as a unit, mark it oft* on AH thirty-two times, beginning at A. 
From I, the last of these dividing points, draw a line to e\ next 
find the points B, C, D, etc., as follows: B is the first division 
from A, C the third from B, etc., thus getting the divisions of 
A1 in the ratio 1, 3, 5, 7, 7, 5, 3, 1. From the points B,* C, D, 




etc., draw lines parallel to le^ meeting Ae at 1, 2, 3, etc., which 
will be the required points of division. 

The cams which we have been considering all act on the com¬ 
paratively sharp end of the piece P, and consequently much fric- 
tion and rapid wearing will result if much power is transmitted. 
Fig. 31) shows a cam designed to act on a roller. The outline is 
first found for a cam to act on a piece like those in the preceding 
cases, with its point at the center of the roller. This curve 
is shown dotted in the figure. Next, with a radius equal to the 
radius of the roller, and with centers at any number of points 
around the dotted curve, draw arcs as shown. A smooth curve 
drawn tangent to these arcs will be the outline of the cam which 
is to act on the roller. 



















MECHANISM. 


39 


Cylindrical Cams. A plate cam, such as we have been con¬ 
sidering, can be designed to give almost any kind of motion in a 
plane at right angles to the axis of the shaft on which the cam is 


located, but can not give motion 
])arallel to the axis of the shaft. 
If, however, we place on the shaft 
a cylinder with a groove cut in 
it, the shape of the groove may he 
made such as to give to a piece 
inserted in it almost any motion 
along the shaft; or, if the depth 
of the groove be varied, the fol¬ 
lower may be given a motion 
which shall be made up of two 
components, one along the shaft 
and one at right angles to the 
shaft. 

FitJ*. dO shows two views of 
a cam designed to give a back¬ 
ward and forward motion in a 
line parallel to the axis of the 
shaft. 

Familiar illustrations of the 
the bobbin-winding mechanism 



use of plate cams are found in 
on some sewing machines; on 



many kinds of looms, and in the valve gear on the Brown and 
other steam engines. Cylindrical cams are used in many places 





















































40 


MECHANTSM. 


ill textile work and in machine tools. One very common use is in 
stretchers on many forms of cloth-finishhig machinery. 


LEVERS. 

A lever is a rigid piece supported at some point called the 
lulcrum, and so arranged that when force is applied at a certain 
point in the ])iece, work is done at a certain other point hy reason 


; FOF 

'' r-J 

1 

FORCE 

WEIGHT 

I 1 

I - 

1 

□ i : 1 

1 ^ FULCRUM 

: 1 

fe—A - B -H 

^ FULCRUM 



WEIGHT 


Fig. 41. Fig. 42. 


of the pivoting action around the fulcrum. Figs. 41, 42 and 43 

illustrate the three common kinds of levers the difference consist- 

/ 

ing in the relative positions of the force applied and the weight 
lifted, with respect to the fulcrum. In any of the three cases, if 
we let F = force applied, AV — weight lifted, A = distance from 
fulcrum to weiglit, and B = distance from fulcrum to F, we have 
the following equation: 


FORCE 




Fulcrum 


F X B = A X AV or 

/AN F ^ 

(i6) 

from which, if any three of the 
quantities are known, the fourth 
may be found. 

Distance A is called the 

weight arm^ and distance B is 

called the The pro¬ 

duct of A times the weight is 

called the moment of the weight, and the product of B times the 
force is called the moment of the force. Another way of stating 
the principle ex])ressed in equation (10j is: 


WEIGHT 

Fig. 43. 


Moment of force — moment of weight. 



































MECHANISM. 


41 


Example: Suppose a weight of 40 pounds is hung on the 
end of a lever. If the distance from the fulcrum to the weight is 
14 inches, what power must be applied to raise the Aveight, the 
iistance from the fulcrum to the power being 18 inches? 

E X -E == A X AC 

E X 18=14 X 40 

.. 14 X 40 

= 31.11 pounds. 

Another example: Let the accompanying diagram represent 
a safety valve. AVe will neglect the weight of the lever and valve. 
Suppose the valve Y has an area of 3 square inches and the steam 
pressure per square inch is 36 pounds. The valve is 5 inches 
from the fulcrum and the weightis 32 inches from the fulcrum. 
AVhat should be the weight of the hall B? 



Eirst, let us hnd the moment of power. Tlie total upward 
j)ressure is 3 X 36 = 108, because the total steam pressure e(pials 
the ])ressure per scpiare inch multiplied by the area. 

The moment of power is 108 X 5 = 540. 

The moment of weight is 32 X B. 

Then 32 X B = 540 
P _ 540 
* 3 ‘^ 

= IT ])ounds (nearly). 

Had we considered the Aveiglit of tlie valve and lever, we 
woaild have added their moments to that of the weiglit, because 
the^e weights increase the weight hut not the ])ower. 

If in addition to the data given al)ove, the lever weighs 10 













42 


MECHANISM. 


])Ouiids and its center of gravity is at its center, that is, Id inches 
from the fulcrum. Then 

Moment of lever = 10 X 16 = 160. 

The valve and valve spindle weigh 8 pounds and are 5 inches 
from the fulcrum, hence 

Moment of valve =8x5 = 40. 

Total moment of weight = 32 X B + 200. 

The moment of power is, from the previous example, 540, 
and we have the ecpiation 

32 X B -f 200 
32 B 
32 B 

B 

Thus we see that by considering the weight of the valve and 
spindle and that of the lever, the ball can be reduced in weight 
from 17 to 10| pounds. 

If the lever is curved as in Fig. 44, the weight arm A becomes 
the length of^the perpendicular from the fulcrum to the line of 
action of the weight, and the power arm B the length of the per¬ 
pendicular from the fulcrum to the line of action of the force. 

LINK WORK. 

Four=Bar Linkage. If we have the four rods AB, BC, CD 
and DA, Fig. 45, each connected to two of the others by pin and 
eye joints, each rod becomes what is called a link, and the whole 
forms a four-bar linkage. If the link AB is fixed in a position 
(as regards the frame of the machine) the pins at A and B become 
fixed centers, the links AD and BC become what are called cranks, 
and the link CD becomes a connecting link or connecting rod. If 
AD is caused to turn with a uniform angular speed about the ])in 
of A, that is, turning through e(pial angles in equal periods of time, 
BC will be comj)elled to turn with some sort of motion about ])in 
B. AVhether i>C shall have uniform ano;ular motion and whether 
it shall make conqdete revolutions for complete revolutions of AD, 
will depend upon the relative lengths of the four links. 

The center line drawn through the fixed link AB is called the 


= 540 

= 540 - 200. 

= 340 
-^ 4-0 

= —^ = 10.62 + pounds. 






MECHANISM. 


43 


line of centers, and the center line drawn through the connecting 
rod CD is called the line of connection. AYliatever the relative 
lengths of the various links, the following law holds true for any 
position of the linkage. 

Angular velocity of crank AD B II BE 
Angular velocity of crank B C A K A E 



That is, if at any given instant the connecting rod be im¬ 
agined removed and the cranks continue turning at the same speed 
at which they were moving when 
the connectino; rod was removed, 
the number of turns of the cranks 
in a given period of time would 
be to each other inversely as the 
lengths of the perpendiculars 
drawn from the respective fixed 
centers to the line of connection, 
and also inversely as the dis¬ 
tances from the lixed centers to 

the point where the line of connection meets the line of centers. A 
mathematical proof can be given for this law, but it will not be 
considered here. 

If the fixed link xVB is made equal in length to the connect¬ 
ing rod DC, and the two cranks are equal (see Tig- T6), the con- 




necting rod will always be parallel to the fixed link, and the per¬ 
pendiculars BII and AK will always be equal, so that from equa¬ 
tion (17) it will be evident that the sj)eeds of the cranks are always 
equal. An examj)le of a linkage of this kind is found in the par¬ 
allel rod connecting the drivers of a locomotive, as shown in Tig. 
47. The axles of the drivers form the fixed centers A and B, the 




















MECIIAMSM. 


• 44 


frame of the locomotive forms the fixed link AB, the drivers them¬ 
selves form the cranks AD and BC, and the parallel rod forms the 
connecting link. 

Crank and Connecting Rod. In the four-bar linkage shown 
in Fig. 45, the pin C is compelled by crank BC to move on the 



Fig. 46. 

arc of a circle whose center is B and radius BC. If in place of 
BC the fixed link is formed as shown in Fig. 48, being provided 
with a slot whose radius is BC and center is B, the pin C on the end 
of the connecting rod will have exactly the same motion that it 
would if guided by the crank BC. How, if the radius of the slot 




hig. 


47. 


is gradually increased, the slot will become nearer and nearer to 
being straight, so that the linkage shown in Fig. 49, where the 
slot is straight, may be said to be the limiting case of the linkage 


shown in Fig. 48. This is the linkage which we have in the crank, 
connecting rod, crosshead and crosshead guides of a steam engine. 
The craidv shaft corresponds to the pin A, the frame and cruides 
































MECHANISM. 


45 


take the.place of the slotted link AC, DC is the connecting rod, 
and AD the crank. 

As there are many problems which may arise in connection 
with this linkage, particularly as applied to the steam engine, we 
will consider the most important. 



Fig. 50 shows the same linkage, the links being represented 
by their center lines only. AD is the crank, DC the connect- 
ino- rod, C the crosshead and AC the center line through the crank 
shaft and the guides. 

Suppose the crank has a length K and is making N revolu¬ 
tions per minute at a uniform speed, to find how fast the crosshead 
is moving at any given position of the crank. 


D 



Me will solve this problem partly graphically and partly by 
calculation. The first step is to find how fast the crank pin D is 
moving. For one turn of the crank, D travels through a distance 
equal to the circumference of a circle whose radius is H, that is, a 
distance of 2rrU. If the crank turns N times per minute the 
velocity of D must be 2 u" liN. Knowing the value of H and A, 
for a-ny particular case we can substite in this formula and find the 
actual velocity of D. Next draw the linkage at a convenient scale, in 
the position in which it is desired to find the velocity of 0. Draw Bd 


























MEORAXTSM. 


4r> 


perpendicular to Ad) and make its lengtli I’epresent at a convenient 
scale (not necessarily the same scale as was used for the links) the 
calculated velocity of 1). This represents the actual velocity and 
•direction of motion of the end 1) of the connectino; rod at the in- 
stant under consideration. Now resolve the velocity V>(1 into its 
components, along and at right angles to DC, by prolonging the 


d 



line DC and drawing <:7E perpendicular to it. The component DE 
is the only one which has any effect on the motion of the point C. 
ATe have already learned that the motion of the two ends of a rod 
must have their components along the rod equal, therefore lay off 
from C, CF — DE. The actual motion of C is aloim the line CA, 
and, as Ch^ is one component of the motion, the other component 



must be perpendicular to CF, therefore draw Fid perpendicular to 
CF and meeting CA at dl. Then Cdl is the velocity of point C 
at the same scale at which J)(l was drawn. 

Eccentric. The eccentric and eccentric-rod linkao-e is in 
principle the same as the crank and connecting rod. To trace out 
the connection between the two, let us consider the crank and con¬ 
necting rod shown in Fig. 51. Here the crank pin D is large and 
comes almost to the center of the shaft A. By increasing still 













IMECHAXISM 


47 


further the diameter of the ])iii 1), so that it will take in the shaft, 
as shown in Fig. 52, we obtain the eccentric. 



pose that in Fig. 55 a force F is a])])lied at the crank pin D in a 
direction perj)endicnlar to the center line AD; then the pressure 
which would be exerted by the slide C could be found bj the same 
])rinciple which we have used in previous cases, namely, the forces 
are inversely proportional to the velocities. Let F be the force at 
D, and P the pressure at C. Assume any velocity of the point D 
and represent this at some scale by the line Dd perpendicular to 
AD. Then find the velocity CII of the point C along the line AC, 
in the same manner as in Fio;. 50. The value of CII can bemeas- 
ured off at the same scale at which IhZ was drawn. Then 


F _ C II 
P “ 1) d 




from which, if one of the quantities F or P is known, the other 
may be found. 

If the value of CII is found for various positions of AD, keep- 
ino- Dd constant, it will be seen that the nearer AD comes to 
coinciding with AC, the smaller CII becomes, and from equation 
(18) it will appear that the smaller CII is, the greater P becomes 
with relation to F. The above discussion will hold true if the 
force acts in the direction D^ or in the opposite direction. 

The linkage used in this way to effect pressure at by a force 
on D is known as the toggle joint^ and is used with more or less 
modification in several kinds of machinery, such as punching ma¬ 
chines, brakes on the drivers of a locomotive, etc. The force need 

















48 


MECKANISM 


not be apj)lied at E>, but may be applied at some other point on 
AD, or on some piece rigidly connected to AD but making an 
angle with it. This is shown in Fig. 54. Here the force is ap¬ 
plied at E, and we can lind the force at C by the same principle 
as before: The force at C is to force at E inversely fis the veloci- 


D 



ties of C and E. The only difference is in finding the velocity of 
(h Assume a velocity for E, as Ec, Then 

Velocity E A E 

A'elocityD AD •'9' 

from which the velocity of D may be found; the rest of the solu- 
tion would be the same as for Fio-. 53. 

O 



These problems may be solved by another and perhaps some¬ 
what shorter method, but, as this would involve the resolution of 
forces, which have not yet been taken up in detail, we will confine 
ourselves to the above method. 

























i\LKC]llANISM. 


49 


QUICK RETURN MOTIONS. 

In various kinds of niacliinery, particularly iiiachine tools, it 
is desired to move a piece backward and forward; the forward 
Tuotion being slow and the return motion more rapid. Take for 
example the shaper; as the tool when moving forward is cutting 
metal, h should go slowly and steadily, but after the cut is made 
it is desirable to get the tool back ready for its next sti’oke as 
(juickly as possible. Many different devices are used to accom¬ 
plish this result. 

If in Fig. 45 we replace the connecting rod DO by a slot in 
the link BO, we get the linkage shown in Fig. 55. A and B are 
the fixed centers, AD is the driving crank (which usually turns 



with uniform speed), BD is the slotted crank and to some part of 
BD a link or some other piece is connected by means of which the 
tool is driven. Two distinct mechanisms are formed, depending 
upon the relative lengths of the links. If the proportions are such 
that a circle drawn around the center A, with radius AD, falls 
outside the center B, as shown in Fig. 50, we have what is known 
as a Whitworth quick return motion. Here the slotted crank 
makes one complete revolution for each complete revolution of 
AD, but its speed is not uniform. In this figure, a connecting 
rod FT is represented as attached to a point B on the slotted link. 
The other end of this connecting rod moves the tool holder T 
along the straight line BT. When the linkage is in the position 
shown, T is in its extreme right-hand position, and it will be in its 













50 


MEOKANTSM. 


extreme left-hand position wlien EP occupies tlie ])osition BPj. 
In turnino BP through tliis anule (ISO^), AD has turned through 
the angle L. In returning BP to its right-hand position again, 
AD has to turn through the angle M only. Now, since AD turns 
with uniform speed and since angle M is less than angle L, T 
makes its stroke from left to right in less time than was required 
to move from right to left. The time of advance and time of 
return are in the ratio of angles L and M. If the length of the 

crank AD and the ratio of time of advance to return are known 
the distance AB may be found as follows: 


/ 

/ 



Fig. 56. 

With A as a center and AD as a radius, draw a circle and 
divide the circumference by the points D and D^ so that angle L 
may bear the same ratio to angle M that the time of advance bears 
to the time of return. Join D and D^ and from A draw a line 
perpendicular to DDj, meeting it at B, which will be the recjuired 
center for the driven crank. 

The distance BP governs the length of the stroke of the tool, 
so that by varying the j)osition of P the length of the stroke may 
be varied. 

If the proportions of the linkage are such that a circle drawn 
with A as a center and a radius AD comes inside the point B, 


















MECHANISM 


51 


tliat is, if AT) is less than All,we liave wliat is known as a swincr- 
in^ block linkage. Tm^- 57 sliows this form of quick return mo¬ 
tion. The ])iece wliicli drives tlie tool is attached to the slotted 
link at P, wliicli may he anywhere alon^ the link. The linkage is 
shown at the extreme right of its stroke, the point J) being at the 
point where the center line of the slot is tangent to the circle 
drawn with center A and radius AD, (that is, the length of the 
driving crank). ADj is the corresponding position of the crank 



Fig. 57. 


at the other extreme of the .stroke. The time of advance is to the 
time of return as anorle L is to angle M, as in the case of the 
Whitworth quick return. If the ratio of advance to return and 
the length of AD are known, the position of B may be found by 
drawing the circle with center A and radius AD, and on this circle 
finding points D and Dj such that the angles L and M shall have 
the required ratio. Then draw DB and D^B tangent to the circle 
at D and Dj respectively, meeting each other at B, which is the 
required center for the driven crank. 

















MECirAxis]\r. 


KO 

O t>j 


GEARING. 

If shaft B, Fig. 58, has fastened to it a disc with smooth cir¬ 
cumference, the disc being in contact at the point P with another 
disc on shaft A, the rotation of one shaft will cause the other to 
rotate, j)rovided there is sufficient friction between the two sur¬ 
faces to prevent slipping. According to the principles which we 
have already learned, the following equation will hold true: 

lievolutions per minute of A BP 
Bevolutions per minute of B A P 

In practice, the friction would not be sufficient to be relied 
upon, so that discs having teeth u])on their circumference are used, 
instead of the plain discs. The discs or wheels thus formed are 
called gears. The teeth may have any one of several forms, and 
the gears may connect shafts which are parallel, intersecting at 
some angle, or neither parallel nor intersecting. Two gears such 

as shown in Fin. 59, where the shafts 
are parallel and the teeth on the cir¬ 
cumference are parallel to the axis 
of the gear, are called spur gears. 

The two dot and dash circles 
drawn about the centers A and B 
and in contact at P correspond to 
the two discs of Fig. 58, and the gears may be said to be 
derived from these circles, which are called the pitch circles. 
The diameters of these circles are called the pitch diameters 
of the gears. The point P, where the two pitch circles touch 
each other, is called the point. Circles drawn through 

the outer ends of the teeth are called the addendum dries.,. 
and circles drawn at the bottom of the teeth are called the 
root circles. These circles are indicated on the figure. That part 
of the tooth outlined between the pitch circle and the addendum 
circle, as PE, is called the face of the tooth, and that j)art between 
the pitch circle and the root circle, as PF, is called the flank. 
The radius of the addendum circle minus the radius of the pitch 
circle is called the addendum distance, or simply the addendum. 
The radius of the pitch circle minus the radius of the root circle 
is called the root distance or the root. The root is commonly 







MK(;J1AN1SM. 


58 


made a little greater than the addendum, so that when two gears 
have their pitch circles in contact, if their teeth are of e(pial length 
they will not touch bottom, l)ut will have some clearance. The 
distance from the center of one tooth to the center of the next, 
measured on the ])itch circle, as (HI, is called the circular pitch, 
or circumferential pitch, and is e(]ual to the circumference of the 
pitch circle divided by the number of teeth. 

In order to run together, two eears must have the same cir- 
cular pitch. The number of teeth on two gears of the same pitch 
are proportional to the circumferences, and, consecpiently, to the 
diameters of their pitch circles; and as the speeds of the two shafts 


t 



which are connected by a pair of gears are inversely proportional 
to the diameters of the pitch circles of the gears, the speeds must also 
be inversely proportional to the number of teeth. 

On rouo-h ^rears the width of the tooth LM is made a little 
less than the width of the space EL, to allow for irregularity of 
construction. The difference in the width of the two is called the 
haah lash. 

Although the circular pitch is a term which is frecpiently 
used in connection with gearing, there is another kind of pitch 
which is often used. This is the diametral pitch, and is e(pial to 
the diameter of the pitch circle divided by the numl)er of teeth, 
or, in other words, the amount of pitch diameter allowed for each 















>4 


*09 







































MECHA^TISM. 


55 



tooth. The diametral ])itch bears the same relation to the circu¬ 
lar ])itch tliat the diameter of a circle hears to its circumference; 
that is, the diametral pitch is e(|ual t(i tlie circular pitch divided 
by TT. In sj)eaking of the ])itc]i of a gear, instead of using the 
diametral ])itch, whicli is often a fraction, it is customary to use 
only the denominator of tlie fraction. For exam])le, if the 
diametral ])itch is 4 inch, it would be spoken of as a 2 -pitch gear. 
This would he the same as saying that for every inch of ])itch* 
diameter there are two teeth. 

Problems similar to the following often arise in this connec¬ 
tion : 


1. Given two gears, 4 pitch, having 12 teeth and 16 teeth 
respectively, to find their pitch diameters and the distance apart 
that their centers should be located, to run properly. 

Since they are 4 pitch, their diametral pitch is ^ inch and 
their pitch diameters will be their numbers of teeth multiplied by 
J, so that one would be 12 X 5 = 3 inches in diameter, and the 
other 16 X 5 = 4 inches in diameter. The distance on centers will 
be the sum of their pitch radii = f + — 34 inches. 

2. Given two gears, 8 pitch, having diameters 6 and 10 
• inches respectively, to find the number of teeth. 

The expression 8 pitch,” besides meaning that the diametral 
])itch is inch, means that there are 8 teeth for every inch of 
diameter, so that one gear will have 


8 X 6 = 48 teeth 
and the other 8 X 10 = 80 teeth. 


Eeferring again to Fig. 59: If gear A is driving the othel* 
gear in the direction indicated, there is a working point of contact 
between the teeth of the two gears at J and another at K, and as 
the gears turn and the teeth slide along each other, the point where 
a ])air of teeth is in contact changes. If from the point of contact, 
as J, a line is drawn to the ])itch P, the tooth curves should have 
such form that this line is normal to both curves at J; that is, PJ 
should be perpendicular to a line drawn tangent to both curves at 
J. This condition should hold, wherever the point of contact may 
be, in order that the gears may run smoothly. 

There are two kinds of curves which are commonly us(‘d for 
the outlines of gear teeth, and which fulfill the above conditions. 


) ) 
> ) > 




6H 


MECHANISM. 















MECHANISM 


57 


_ • 

These are the involute and the epicycloidal curves. Eig. 60 
shows a pair of gears with epicycloidal teeth and Fig. 61 shows 
parts of two gears with involute teeth. The construction for get¬ 
ting the tooth outlines is shown in the figures, but as this properly 
comes under the subject of drawing, and is not essential to an 
understanding of gears in general, it will not be explained here. 



Annular Gears. An annular gear is a ring with teeth on its 
inside edge. Fig. 62 shows such a gear with center at A, mesh¬ 
ing with a small spur gear called the pinion. 

Rack and Pinion. A rack is a gear whose pitch line is a 
straight line instead of a circle. Fig. 63 shows an epicycloidal 
rack in gear with a pinion. 

































58 


MECHANISM 
























































MECITANISM. 


50 


Bevel Gears. AVlieii two sliafts wliose axes intersect are to 
be connected ])y ^ears, the ^ears coninionly nsed are called bevel 
gears. The theoretically correct ])itcli surface of a bevel gear is a 
])art of the sector of a S|)bei*e, but as the designing and making 
of sncli a gear is somewhat dithcult, a cone is snbstitnted for a 
sphere. Eig. (>4 shows a])air of bevel gears connecting two shafts 



Fig. 64. 


- n 


which are at right angles. O is the ])oint where the center lines 
of the shaft intersect, OPA and OPB are the 2 ^itch cones of the 
rears, AP and BP the pitch diameters. 

The speed of the shaft S is to the speed of the shaft T as AP 
is to BP. The teeth may be epicycloidal or involute, and the 
shafts may be at right angles or at some other angle. 

Trains of Wheels. Thus far both in gearing and in pulleys 
and belts, we have considered only one ])air of wheels, one being 
on the driving shaft and the other on the driven shaft. It veiy 







































00 


MECHANISM. 


often happens that the connection cannot he made directly by one 
pair of wheels, either because the shafts are not conveniently ])laced 
with relation to each other, or because the diameters cannot be so 
proportioned as to ^ive the desired speed. In such cases the con¬ 
nection may be made by a train of pulleys or a train of ^ears. 
Fig. 05 shows a train of pulleys, and Fig. 00 a train of gears. 

In Fig.'H)5 shaft A drives shaft E by means of judleys Ill and 
F. E in turn drives C by means of ])ulleys G and IJ, and C drives 
I) by pulleys J and K. 

^Vssunie shaft A to make 100 li. P. M. and the pulleys of the 


following diameters; 

O 


E = 30 inches G = 21 inches J = 20 inches 

E = 12 inches II = 18 inches X = 10 inches, 

to lind the number of revolutions of shaft 1). Applying the prin¬ 
ciples which we have already learned, 

E. P. M. of E Diameter of E 

- ■—: - nr 

11. P. M. of A Diameter of F 

K. r. M. of B = K. P. M. of A X = loo x 41 

diam. of 


Ill like manner K- P. M. of C = E. P. j\I. of B X yI 

= 100 X 41 X -ih 

and 11. P. M. of D = K. P. M. of C X fj- 

= 100 X fi X u X 41 
= 800. 

\ 

It will be noticed that pulleys E, G and J are drivers and F, 
II and K are driven pulleys, so that the above maybe stated thus: 

Eevolutions of last shaft (or last wheel) = revolutions of first 
shaft (or first wheel) multiplied by a fraction whose numerator is 
the ])roduct of the diameter of all the (Irivlmj pulleys, and whose 
denominator is the product of the diameters of all the driven 
pulleys. (20) 

The same rule will apply to the train of gears in Eig. (16, if. 
we use either the pitch diameters or the number of teeth in tlie 
respective gears where diameters are used instead of the j)ulley 
diameters. 

In Fig. 05, where all the belts are open belts, all the shafts 









MECHANISM. 


61 


turn in tlie same direction; but where two shafts are connected by 
a j)air of sjmr gears, tlie sliafts turn in oj)j)osite directions, so that, 
in Fig. 66, the direction in which shaft I) will turn with reference 
to the direction of A, will depend upon the number of gears in the 
train. The best way to determine this is to follow the train 
through, ])utting arrows, as shown, on the res])ective shafts. 



Idle Gear. In Eio-. (>7, gear E drives gear F, and F drives 

O “ o' 

(t, that is, F is both a driven and driving gear. Let K have 25 
teeth, F 30 teeth, and G 50 teeth, and let shaft C make 100 R. P. 
l\r. Then from the rule given above, 

K.P. M. of A =100 X 15 X ill . 

In this equation the number of teeth of F (30) occurs both in 



the numerator and denominator of the fraction, and will therefore 
cancel each other, so that the equation becomes 

IL P. M. of A = 100 X 

Therefore, the speed of A depends only upon the number of teeth 
in E and G. Gear F is called an idle gear or idler. It should be 
noticed, however, that A will turn in the oj)posite direction from 
wliat it would if G geared directly with E. 

In any train of wheels, the number of turns of the last wheel 


























02 


MEdirANTSM. 


(as K ill Fi^s. (>5 and (>(>, or (I in Fig. (iT) divided liy tlie mindier 
of turns of tlie iirst wheel, as F, is called valuk of the train. 
(\)iisidering this statement in connection witli formula (20) we 
have the following:— 


Value of train = 


Product of teeth of all drivers 
Product of teeth of all driven wheels. 



This value will be jiositive or negative, that is, it will have the 
aWehraic siorn + or — before it, accordintr as the last wheel turns 
in the same direction as the hrst, or in the opposite direction. 

Epicyclic Trains. In Fig. 08 the gear B is loose on the shaft 
11 and driven by the gear F; arm A is keyed to the shaft II so 
that it turns when the shaft turns. Gears 0 and D are carried on 
studs on the arm. It will lie evident that if B and If are caused 



to turn independently, the gear D wdll have a different number of 
turns on its own axis from what it would if A were fixed and 1) 
were dilven directly through the train BCD. 

In the following formula, the value of the train is the ratio of 
the turns of D to the turns of B if the arm were fixed, consider¬ 
ing B the first wheel and I) the last; by the turns of B and 1) 
respectively we mean the actual number of turns they make when 
the arm is in motion. The formula by which the turns of A, B, 
or 1) can be found is as follows,— 


. d urns of last wheel 

\ alue ot train = fp, - ^^—r 

i urns ot lirst wheel 


turns of arm 
turns of arm 



This can be easily jiroved, but the jiroof need not be studied 
to understand the use of the formula. In applying this, the proper 
algebraic signs must be given to the value of the train, and the 























MKOirAmSM. 


03 


turns of tlie arm: That is, if witli tlie arm tixod, B and I) would 
turn ill the same direetion, tlie value of the train is eonsideriMl 
jdus, if in ojijiosite directions, the value of the train is minus. If 
the arm turns in the same direetion as H, the turns of the arm 
would be jilns, but if in the opposite direction the turns of the 
arm would be minus, so that the formula would become, 


Value of train = 


Turns of last wheel + turns of arm 
Turns of first wheel + turns of arm 


That is, two minus signs before the number of turns of arm would 
make it plus. This may be more clearly understood from a study 
of the following problems: 

In Fig. 68 let B have 75 teeth, C) 30 teeth and I) 25 teeth, 
and let B be driven by E at a speed of 20 B. P. M. right-handed, 



while A is driven at a speed of 2 B. P. M left-handed, to find the 
number of B. P. M. of I) and the direction in which it will turn. 
AVe will use formula (22). The value of the train will be found 
from formula (21). 


Value of train = 


Teeth of B X Teeth of C 
Teeth of C X Teeth of I) 


Value of train = X fy = 3 

and since with the arm fixed B and T) would turn in the same 
direction, the value of the train will be ])lus. Since the arm A 
turns in the opposite direction from B, its number of turns will 
have a minus sign. Then substituting the formula (22.) 

B. P. ]\r. of D — (-2) B. P. M. of D V 2 




























04 


MECITAXTSM. 


or 

3 X 22 == K. P. M. of 1) -f 2 
whence 64 — K. P. M. of I) 



Tn Fig. 69 the gear B, the arm A, and the annular gear D 

liave a common axis, hut are 
arranged to turn independ¬ 
ently of each other about this 
axis. C is an idle wheel con¬ 
nectings B with the annular 
gear D and is on an axis car¬ 
ried by A. Let B have 30 
teeth, C 45 teeth and I) 120 
teeth, and let B be caused to 
have 125 B. P. Al. right- 
handed and arm A have 75 
K. P. M. left-handed, to find 
the number of turns and di¬ 
rection of motion of D, First 
find the value of the train if A were fixed, which would be 


(30 X 45) ^ , 
(45 X 120) ^ 



ttience substituting in formula (22). 

^ Turns of D — (-"<^5) Turns of D + 75 
~ 125 — (-75) "" 

whence turns of I) = — 125 


200 

































































MECHANISM. 


65 


tliat is D, makes 125 K. P. M. in the opposite direction from that 
in which B is turn in or. 

In Fig. 70, gears B and D have a common axis, B is fast to 
the frame of a machine, D is fast to the shaft S, which drives the 
work on the machine, A is a disc centered on the hub which snp- 
])orts B, and driven by a gear (not shown) which meshes with a 
gear on the back of A. C is a gear carried on the stud in A. Tlie 
gear B corresponds to the first wheel or driver of an epicyclic train, 
A to the arm, C is an idle wheel, and D is the last wheel of the 
train. B has one tooth more than D. Let B have 35 teeth, C 15 
teeth, D 31 teeth, and let A make one turn right-handed, to find 
the speed of D and the direction of its motion. The value of the 
train, considering A fixed and B the driver, is |^|. Then formula (22). 

35 Turns of D — 1 

31 0 — 1 

or — 35 = 31 Turns of T) — 31 
— 1 =31 Turns of I) 



or Turns of D = — That is, for one turn of A right-handed, 
with B fixed, D makes of a turn left-handed. 

In Fig. 71 is shown an .epicyclic train of bevel gears. The 
shaft S carries the shaft A either welded to it or clamped on 
it in some way. Loose on S are the bevel gears B and D, held in 
])lace by collars (not shown) and loose on A is the bevel C, gear¬ 
ing with each of the others. Fast to B is the spur gear E, which 
is driven from some source of power not shown. S is driven from 





































68 


MECHANISM. 


some external source of power, thus causing the arm A to swing 
around. Let all of the bevels have the same number of teeth, and 
let B be driven at a speed of 50 11. P. M. right-handed, and S at 
a speed of 25 R. P. M. right-handed, to find the speed and direc¬ 
tion of D. Formula (22) applies here as in other cases which we 
have considered. 


Then 


The value of the train 
Turns of D 


25 


50 — 25 
— 25 = Turns of D — 25. 

Therefore, Turns of D = 0. 

Again let B turn 50 B. P. M. right-handed, to find the direc¬ 
tion and number of turns of A in order that D shall turn 25 
R. P. M. left-handed. 

— 25 —Turns of A 


— 1 


50 — Turns of A 


— 50 + Turns of A = — 25 — Turns of A. 

Therefore, Turns of A = 12J right-handed. 






EXAMINATION PAPER 


» V 




MECHANISM. 


Read carefully: Place your name and full address at the head of the 
paper. Any cheap light paper like the sample previously sent you may be 
used. Do not crowd your work, but arrange it neatly and legibly. Do not 
cojiy the answers from the Instruction Paper: use your own ivords, so that 
ire may he sure that you understand the subject. After completing the work 
add and sign the following statement: 

I hereby certify that the above work is entirely my own. 

(Signed) 


1. An engine makes 280 revolutions per minute. If the 
stroke is 16 inches, what is the linear velocity of the crank pin in 
feet per second? 

2. The pitch of a jack screw is ^ inch. The length of the 
bar is 4 feet. What weight can be lifted if the force applied at 
the end of the bar is 148 pounds? 

3. A worm of double thread meshes with a worm wheel 
having 64 teeth. IIow many revolutions per minute must the 
worm have to turn the wheel 2 times per minute? 

4. A lever safety valve has a lever 30 inches long. The 
weight hung on the end weighs 24 pounds. If the valve is 44 
inches from the fulcrum and has an area of 3^ scpiare inches, at 
what pressure per square inch will it blow oft* ? 

IN^eglect weight of valve, spindle and lever. 

5. Define circular pitch. Diametral pitch. 

6. If a gear has 20 teeth and it is 6 pitch, how far will 
the rack move in 24 revolutions of the gear? 

7. PTnd the piston speed in feet per minute of an engine 
making 5.4 revolutions per second, the stroke being 18 inches. 

8. Two gears mesh with each other. They are 8 pitch. 
If they have 40 teeth and 68 teeth, respectively, what are the 
diameters and how far apart should the centers of the shafts be 
placed? 

9. The pitch circle of an annular gear is 36 inches in diam¬ 
eter. The spur gear meshing with it is 8 pitch and has 24 
teeth. How many revolutions must the spur gear make to turn 
the annular gear 13 times per minute? 




MECHANISM. 


10. A lever is 26 inclies long. A weight of 20 pounds is 
hung on one end. The distance of this weight from the fulcrum 
is 11 inches. What power will be necessary to raise the weight 
if the distance from the fulcrum to the power is 15 inches ? 

11. Describe the eccentric‘and eccentric-rod linkage. 

12. A worm of single thread is in mesh with a worm wheel 
havinpf 36 teeth. A crank of 16 inches radius is attached to the 

O 

worm and a drum 10 inches in diameter is fastened to the worm 
wheel. How fast will the weight rise if the crank is turned 40 
times per minute? 

13. Suppose the press shown in Fig. 32, page 31, has a hand 
wheel 16 inches in diameter. The pitch P = ^ inch and pitch p 
— ^ inch. If 80,000 pounds pressure is to be exerted, what force 
must be applied to the hand wheel? 

14. Describe some form of friction clutch. 



15. In the train of gears shown, A is attached to the shaft 
of a motor and makes 1,000 revolutions per minute. How many 
revolutions per minute will gear D make? 

16. If the linear velocity of the rim of a fly wheel is limited 
to a mile a minute, how many revolutions are allowable if the fly 
wheel is feet in diameter? 

17. A locomotive is running at the rate of 50 miles per hour. 
The drivers are 6 feet in diameter and the linear velocity of the 
crank pin is 24.44 feet per second. What is the length of stroke? 

Note. The stroke=the diameter of crank-pin circle, 

18. What is the pitch of a screw that advances 3 inches in 
24 revolutions? 

19. What weight can be raised by the device described in 
question 12 if the force applied to the crank is 110 pounds? 







MECHANISM. 


20. In the accompanying sketch, the large stone weighs 1,248 
pounds. AVhat force must be exerted at the end of the lever to 
move the stone? 

NoTEi Assume that 4J per cout of the force necessary to lift the stone will move it. 



21. If a gear has 48 teeth and the diameter of the pitch circle 
is IG inches, what is the circular pitch? What is the diametral 
pitch? 

22. A gear has IG teeth and runs 800 revolutions per minute. 
How fast will a gear meshing with it run if it has 5G teeth? 

23. The two sprocket wheels of a bicycle have 8 and 24 teeth, 
respectively. If the wheels are 28 inches in diameter, how many 
times must the pedals turn while the bicycle is going, one mile? 

24. In the above example, how many turns per minute must 
the pedals make if the bicycle moves at a velocity of a mile in 58 
seconds? 

25. What is pitch diameter? 

2G. If motion is desired along a shaft as well as at right 
angles to a shaft, what kind of cam must be used? 

27. Two pulleys are connected by a leather belt. The driver 
is keyed to a shaft making 80 revolutions per minute, and is 34 
inches in diameter. How many revolutions will the driven pulley 
make if it is 14 inches in diameter? 

28. To what class does the lever of the lever safety valve 

belonn;? 







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